Find $x$ angle in triangle I need to find angle x in this isosceles triangle(20-80-80), by using pure geometry, if i can say so. If my calculations are correct (i tried another approach) answer should be 30, but there should be 'easy' way to find this.
Also i found many Langley’s Adventitious Angles exercises which are very similar to mine but yet different.

 A: And my second solution is as follow:

A: Let in $\Delta ABC$ we have $AB=AC$, $\measuredangle A=20^{\circ}$ and $\measuredangle ADC=x$ as on your picture. 
Let $M\in AB$ such that $AD=MD$ and $K\in DC$ such that $MK=AD$.
Also, let $B'\in MB$ such that $MB'=AD$ and $C'\in KC$ such that $B'C'||BC.$
Thus, $$\measuredangle MKA=\measuredangle MDK=2\cdot20^{\circ}=40^{\circ}$$ and from here
$$\measuredangle B'MK=40^{\circ}+20^{\circ}=60^{\circ},$$ which says $$B'K=MB'=AD=BC.$$
But $$\measuredangle B'KC'=60^{\circ}+20^{\circ}=80^{\circ}=\measuredangle BCA=\measuredangle B'C'A.$$
Thus, $$B'C'=B'K=AD=BC,$$ which says that
$$B\equiv B'$$ and $$C\equiv C'.$$
Id est, 
 $$\measuredangle BDC=10^{\circ}+20^{\circ}=30^{\circ}.$$
A: 
construct triangle $\Delta BCE$ congruent to $\Delta ADB$.
so $AB = BE$, $\angle ABE = 80° - 20°  = 60°$ 
Thus triangle $\Delta ABE$ is equilateral.
$AB = AE = AC$, since $\angle CAE = 60° - 20° =40°$ 
$\angle AEC = \frac{180° - 40°}{2} = 70°$ 
so $x = 20° + \angle ABD = 20° + \angle CBE = 20°  + (70° - 60° ) = 30°$ 
A: Construct an equilateral triangle such that its sides are equal to the base of the main triangle.

